I think that the effect of playing Best-of-3s and Best-of-5s is often overestimated. If you crunch then numbers you will find that:

When player X has a 60% chance of beating player Y in each game then (ignoring stamina and psychological issues) the probability of X beating Y in a Best-of-3 rises to only 65% and a Best-of-5 pushes the figure up a bit more to 68%.

If instead of 60% you put in 70%, then Best-of-3 gives 78% and Best-of-5 84%.

So, given a clear superiority of one player over another it is by no means certain that a Best-of-N match will be won by the better player.

On the other hand, more challenging conditions will help to nullify the 'luck' element, notably one good hit-in in easy conditions resulting in an all-round break, a missed lift and subsequently a triple peel win. So, if Player X has a 60% chance in easy conditions the probability of his winning each game might rise to, say, 70% in challenging conditions.

Or, to put it another way, a single game win at Surbiton [difficult conditions] could be worth more than a Best-of-5 at Hurlingham ...

Sam Murray supplied a graph showing the probability of match wins as a function of game win %, and number of games in match:

Diagram created by Teodora Baeva

Joel Taylor gives the background to the calculation

Let us assume player A has a probability of p of winning any given match. Thus, A’s opponent, B has a probability of q = (1 − p) of winning each game.

Best-of-3 Games

The probability of player A winning a Best-of-3 match can be determined by considering the total number of games as different cases. A can win (W) with the following score lines:

Score

Probability

=

W-W

p * p

p2

W-L-W

p * q * p

p2q

L-W-W

q * p * p

p2q

Probability of A winning =

p2 + p2q + p2q

=

p2 [ 1 + 2q ]

The probability of player A winning the Best-of-3 match is therefore: p2 [ 1 + 2q ]

The 2 comes from the binomial distribution of the first two games. This idea can be extended to Best-of-5 games.

Best-of-5 Games

The same analysis for the Best-of-5 games gives:

Score

Probability

=

W-W-W

p * p * p

p3

L-W-W-W

q * p * p * p

p3q

W-L-W-W

p * q * p * p

p3q

W-W-L-W

p * p * p * q

p3q

L-L-W-W-W

q * q * p * p * p

p3q2

L-W-L-W-W

q * p * q * p * p

p3q2

L-W-W-L-W

q * p * p * q * p

p3q2

W-L-L-W-W

p * q * q * p * p

p3q2

W-L-W-L-W

p * q * p * q * p

p3q2

W-W-L-L-W

p * p * q * q * p

p3q2

Probability of A winning =

p3 + 3p3q + 6p3q2

=

p3 [ 1 + 3q +6q2 ]

=

p3 [ 1 + (3C2) q + (4C2) q2 ]

So the probability of player A wining the Best-of-5 match is the sum of the individual probabilities.

We use the shorthand nCk for Binomial Coefficients, which corresponds to n!/k!(n-k)!, and is the number of ways of picking k unordered outcomes from n possibilities. E.g. 3C2 is the number of ways you can pick 2 objects out of three.

nCk

n!/k!(n-k)!

=

2C2

2 / 2*1

1

2C1

2 / 1

2

3C2

6 / 2*1

3

4C2

24 / 2*2

6

Best-of-n Games

In a Best-of-n games match (where n is odd), to win, a player must win ½(n + 1) games.

So the probability of winning in straight games is p½ (n+1). Following the pattern of before, the probability of winning in any other number of games and thus the total probability of player A winning is:

Comments

Samir Patel adds:

I’d offer the comment that “it is by no means certain that a Best-of-N match will usually [now redacted] be won by the better player” is a wrong statement.

For two reasons:
1. It is certain that it will *usually* be won by the better player. The graph proves that quite clearly. But it will not always be won by the better player. That’s part of the point of sport.
2. I think it also misses the point that what you’re trying to do is find the better player *on the day*, whilst also reducing randomness on the day (the hill, gust of wind, just snicked rather than just missed roquet…).